12、Anagrams by Stack

How can anagrams result from sequences of stack operations? There are two sequences of stack operators which can convert TROT to TORT:

[
i i i i o o o o
i o i i o o i o
]

where i stands for Push and o stands for Pop. Your program should, given pairs of words produce sequences of stack operations which convert the first word to the second.

Input

The input will consist of several lines of input. The first line of each pair of input lines is to be considered as a source word (which does not include the end-of-line character). The second line (again, not including the end-of-line character) of each pair is a target word. The end of input is marked by end of file.

Output

For each input pair, your program should produce a sorted list of valid sequences of i and o which produce the target word from the source word. Each list should be delimited by

[
]

and the sequences should be printed in "dictionary order". Within each sequence, each i and o is followed by a single space and each sequence is terminated by a new line.

Process

A stack is a data storage and retrieval structure permitting two operations:

• Push - to insert an item and
  Pop - to retrieve the most recently pushed item

We will use the symbol i (in) for push and o (out) for pop operations for an initially empty stack of characters. Given an input word, some sequences of push and pop operations are valid in that every character of the word is both pushed and popped, and furthermore, no attempt is ever made to pop the empty stack. For example, if the word FOO is input, then the sequence:

• i i o i o o
  is valid, but
  i i o
  is not (it's too short), neither is
  i i o o o i
  (there's an illegal pop of an empty stack)

Valid sequences yield rearrangements of the letters in an input word. For example, the input word FOO and the sequence i i o i o o produce the anagram OOF. So also would the sequence i i i o o o. You are to write a program to input pairs of words and output all the valid sequences of i and o which will produce the second member of each pair from the first.

Sample Input

madam
adamm
bahama
bahama
long
short
eric
rice

Sample Output

[
i i i i o o o i o o 
i i i i o o o o i o 
i i o i o i o i o o 
i i o i o i o o i o 
]
[
i o i i i o o i i o o o 
i o i i i o o o i o i o 
i o i o i o i i i o o o 
i o i o i o i o i o i o 
]
[
]
[
i i o i o i o o 

]

 

解题思路：这道题我拖了很久，一是因为准备CCF和网络四级，还有就是因为C++的很多模板都不了解。对于这道题，很明显需要将入栈出栈所有可能性都搜索出来，适合用DFS做。DFS的一般套路为：在开始部分有一个判断是否满足所需条件的语句，接下来就是dfs的递归调用，调用之后需要将操作还原，以便下一次搜索。这道题用到的C++模板有stack和vector，stack用于字符串的存放及其出入栈，vector用于i o的存放及其出入栈。关于vector的更多用法参考http://www.cnblogs.com/wang7/archive/2012/04/27/2474138.html

 

具体代码：

#include<iostream>
#include<stack>
#include<vector>
#include<string>
using namespace std;
string a, b;
stack<char> real;
vector<char> surface;
int length;

void dfs(int numpush, int numpop){
     if (numpush == length&&numpop == length){
           for (int i = 0; i < 2*length; i++){
                cout << surface[i] << " ";
           }
           cout << endl;
     }
     if (numpush < length){
           real.push(a[numpush]);
           surface.push_back('i');
           dfs(numpush + 1, numpop);
           real.pop();
           surface.pop_back();
     }
     if (numpop < numpush&&numpop < length&&real.top() == b[numpop]){
           char temp = real.top();
           real.pop();
           surface.push_back('o');
           dfs(numpush, numpop + 1);
           real.push(temp);
           surface.pop_back();
     }
}

int main(){
     while (cin >> a >> b){
           length = a.length();
           cout << "[" << endl;
           dfs(0, 0);
           cout << "]" << endl;
     }
     return 0;
}

做题感悟：其实对dfs的使用还不是很好，比如在CCF上有一道题感觉是用dfs做，但是因为不知如何控制条件最后也没做出来。药不能停，要坚持练习。